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```Chapter 7: Variation in repeated samples – Sampling distributions
Recall that a major objective of statistics is to make inferences about a
population from an analysis of information contained in sample data.
Typically, we are interested in learning about some numerical feature of
the population, such as
• the proportion possessing a stated characteristic;
• the mean and the standard deviation.
A numerical feature of a population is called a parameter.
The true value of a parameter is unknown. An appropriate sample-based
quantity is our source about the value of a parameter.
A statistic is a numerical valued function of the sample observations.
Sample mean is an example of a statistic.
The sampling distribution of a statistic
Three important points about a statistic:
• the numerical value of a statistic cannot be expected to give us the
exact value of the parameter;
• the observed value of a statistic depends on the particular sample that
happens to be selected;
• there will be some variability in the values of a statistic over different
occasions of sampling.
Because any statistic varies from sample to sample, it is a random
variable and has its own probability distribution.
The probability distribution of a statistic is called its sampling
distribution.
Often we simply say the distribution of a statistic.
Distribution of the sample mean
Statistical inference about the population mean is of prime practical
mean and its sampling distribution.
An example illustrating the central limit
theorem
Figure 7.4 (p. 275)
Distributions of X for n = 3 and n = 10 in sampling from an asymmetric population.
Example on probability calculations
for the sample mean
Consider a population with mean 82 and standard deviation 12.
If a random sample of size 64 is selected, what is the probability that the sample mean
will lie between 80.8 and 83.2?
Solution: We have μ = 82 and σ = 12. Since n = 64 is large, the central limit theorem
tells us that the distribution of the sample mean is approximately normal with
E ( X )    82 , sd ( X ) 

n

12
 1 .5
64
Converting to the standard normal variable:
Z 
X 

n

X  82
1 .5
Thus,
P [ 80 . 8  X  83 . 2 ]
 P [( 80 . 8  82 ) / 1 . 5  Z  ( 83 . 2  82 ) / 1 . 5 ]
 P [  . 8  Z  . 8 ]  . 7881  . 2119  . 5762
```
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